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(B) For parallel first-order reactions,the effective rate constant $k_{eff}$ is the sum of individual rate constants: $k_{eff} = k_1 + k_2$. Since $k

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$2$

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(B) For parallel first-order reactions,the effective rate constant $k_{eff}$ is the sum of individual rate constants: $k_{eff} = k_1 + k_2$. Since $k = \frac{\ln 2}{t_{1/2}}$,we have $\frac{\ln 2}{t_{eff}} = \frac{\ln 2}{t_1} + \frac{\ln 2}{t_2}$. This simplifies to $\frac{1}{t_{eff}} = \frac{1}{t_1} + \frac{1}{t_2}$. Given $t_1 = 12 \ min$ and $t_2 = 3 \ min$,we calculate $\frac{1}{t_{eff}} = \frac{1}{12} + \frac{1}{3} = \frac{1+4}{12} = \frac{5}{12}$. Thus,$t_{eff} = \frac{12}{5} \ min = 2.4 \ min$. The nearest integer is $2 \ min$.

$A$ molecule undergoes two independent first order reactions whose respective half-lives are $12 \ min$ and $3 \ min$. If both the reactions are occurring simultaneously,the time taken for the $50 \%$ consumption of the reactant is $.......... \ min$ (Nearest integer).

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